A ball is thrown froward horizontally from the top of a cliffwith a velocity of
ID: 1722121 • Letter: A
Question
A ball is thrown froward horizontally from the top of a cliffwith a velocity of 10 m/s. the height of the cliff is 45 m. What isthe time taken for the ball to reach the ground?The distance from the cliff of the ball to the horizontal justbefore it hits the ground? And what is the direction of the ball to the horizontal justbefore it hits the ground? A ball is thrown froward horizontally from the top of a cliffwith a velocity of 10 m/s. the height of the cliff is 45 m. What isthe time taken for the ball to reach the ground?
The distance from the cliff of the ball to the horizontal justbefore it hits the ground? And what is the direction of the ball to the horizontal justbefore it hits the ground?
Explanation / Answer
Given that the initial height is h = 45 m initial horizontal velocity is u = 10 m/s ------------------------------------------------- The vertical distance covered is h = uy*t+(1/2)gt2 45m = 0 + (1/2)(9.8m/s2)t2 ( since initial verticalvelocity is uy = 0 ) then we get t = 3.03 s The horizontal distance covered by the ball is R = horizontalvelocity *time of flight = u *t = 30.3 m The final vertical velocity is Vy = Uy + gt = 0 + (9.8m/s2)*3.03s = 29.7 m/s Then the angle with horizontal is =tan-1(Vy / u) =71.4o below horizontal
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